Difference between revisions of "2021 Fall AMC 10A Problems/Problem 25"
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The distance from a point <math>(x,y)</math> to a line <math>y = mx+c \implies mx-y+c = 0</math> is <math>\frac{|mx-y+c|}{\sqrt{m^2+1}}</math>. | The distance from a point <math>(x,y)</math> to a line <math>y = mx+c \implies mx-y+c = 0</math> is <math>\frac{|mx-y+c|}{\sqrt{m^2+1}}</math>. | ||
− | We can set these two equal to each other and we get: <cmath> | + | We can set these two equal to each other and we get: <cmath>\sqrt{(x-a)^2+(y-b)^2} = \frac{|mx-y+c|}{\sqrt{m^2+1}}</cmath> |
+ | |||
+ | Squaring both sides of the equation, we get <cmath>(x-a)^2+(y-b)^2 = \frac{(mx-y+c)^2}{m^2+1}</cmath>. | ||
+ | |||
+ | Expanding both sides of the equation gives <cmath>x^2-2ax+a^2+y^2-2by+b^2 = \frac{m^2x^2+y^2+c^2+2mxc-2yc-2mxy}{m^2+1}</cmath> | ||
+ | |||
+ | Multiplying both sides of the equation by <math>m^2+1</math> and rearranging gives <cmath>x^2+2mx+m^2y^2-2((m^2+1)*a + mc)x-2((m^2+1)*b - c)y +(m^2+1)(a^2+b^2)-c^2</cmath> | ||
Revision as of 23:27, 22 November 2021
Contents
Problem
A quadratic polynomial with real coefficients and leading coefficient is called
if the equation
is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial
for which the sum of the roots is maximized. What is
?
Solution 1
Let and
be the roots of
. Then,
. The solutions to
is the union of the solutions to
and
. It follows that one of these two quadratics has one solution (a double root) and the other has two. WLOG, assume that the quadratic with one root is
. Then, the discriminant is
, so
. Thus,
, but for
to have two solutions, it must be the case that
*. It follows that the sum of the roots of
is
, whose maximum value occurs when
. Solving for
yields
. Therefore,
, so
.
For
to have two solutions, the discriminant
must be positive. From here, we get that
, so
. Hence,
is negative, so
.
~ Leo.Euler
Solution 2 (Factored form)
The disrespectful function has leading coefficient 1, so it can be written in factored form as
. Now the problem states that all
must satisfy
. Plugging our form in, we get:
The roots of this equation are
. By the fundamental theorem of algebra, each root must have two roots for a total of four possible values of x yet the problem states that this equation is satisfied by three values of x. Therefore one equation must give a double root. Without loss of generality, let the equation
be the equation that produces the double root. Expanding gives
. We know that if there is a double root to this equation, the discriminant must be equal to zero, so
.
From here two solutions can progress.
Solution 2.1 (Fastest)
We can rewrite as
. Let's keep our eyes on the ball; we want to find the disrespectful quadratic that maximizes the sum of the roots, which is
. Let this be equal to a new variable,
, so that our problem is reduced to maximizing this variable.
We can rewrite our equation in terms of m as
.
This is a quadratic in m, so we can use the quadratic formula:
It'll be easier to think without the square root, so let : then we can rewrite the equation as
. We want to maximize m, so we take the plus value of the right-hand-side of the equation. Then;
To maximize m, we find the vertex of the right-hand side of the equation. The vertex of is the average of the roots of the equation which is
. This means that since
,
.
.
Solution 2.2 (Derivation-Rotated Conics)
We see that the equation is in the form of the general equation of a rotated conic -
. Because
, this rotated conic is a parabola.
The definition of a parabola is the locus of all points that are equidistant from a point (focus) and line (directrix). Let the focus and directrix of this particular parabola be and
. Then we can try to find the general form of a rotated parabola in terms of
and
.
The distance between two points and
is
. Therefore this is the distance from any point on the parabola to the focus.
The distance from a point to a line
is
.
We can set these two equal to each other and we get:
Squaring both sides of the equation, we get .
Expanding both sides of the equation gives
Multiplying both sides of the equation by and rearranging gives
Solution in progress
~KingRavi